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# A set consist of 2n-1 element. What is the number of subsets

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Manager
Joined: 03 Feb 2010
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A set consist of 2n-1 element. What is the number of subsets [#permalink]

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22 Apr 2010, 12:37
00:00

Difficulty:

55% (hard)

Question Stats:

54% (02:11) correct 46% (01:56) wrong based on 24 sessions

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A set consist of 2n-1 element. What is the number of subsets of this set which contain at most n-1 elements?

A. 2^(2n-2)
B. 2^(2n) - 2
C. 2^(2n) -1
D. 2^(2n)
E. 2^(2n-1)

I dont really know what this question is asking. I dont know the reasoning or the idea behind an empty subset. Can someone explain this question and answer as if they were explaining it to a beginner? Thank you.
[Reveal] Spoiler: OA
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Kudos [?]: 1567 [0], given: 235

Re: A set consist of 2n-1 element from Walker's collection [#permalink]

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22 Apr 2010, 13:01
it is saying if you have some elements a1,a2 till 2n-1 terms and you have to choose any number of items from it and form another set(which you can say is subset), in how many ways you can form the subset if you can choose at most n-1 terms..

First of all please tell the source as these type of questions seems to be out of scope.

IMO ans should be A

I took n=2 so total terms becomes 2n-1 = 3 and at most you can take n-1 = 1 term.

so the ans should be 3, but if you put n=2 in all the equations you wont get the ans. That means they have considered empty set as well, thus the total number of subsets with at most one is a1,a2,a3,0
so ans is 4.

put n=2 in all the equations , you will get A=4 hence A.
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Re: A set consist of 2n-1 element from Walker's collection [#permalink]

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29 Dec 2013, 17:31
1
KUDOS
ksharma12 wrote:
A set consist of 2n-1 element. What is the number of subsets of this set which contain at most n-1 elements?

A. 2^(2n-2)
B. 2^(2n) - 2
C. 2^(2n) -1
D. 2^(2n)
E. 2^(2n-1)

I dont really know what this question is asking. I dont know the reasoning or the idea behind an empty subset. Can someone explain this question and answer as if they were explaining it to a beginner? Thank you.

Yes answer is A indeed. Please allow me to show my procedure

I used n=3, so then we have

5!/2!3! + 5!/4!1! + 5!/0!5!

10 + 5 + 1 = 16

So our target is 16

A gives us 2^4 = 16

Hence A is the correct option

Read carefully it says at most so keep in mind that picking a small number such as 3 will help you save time since you have to list fewer outcomes
Avoid 2 since you will get 1 arrangement (n-1) and may be risky since 1 is a number with certain unique properties

Hope all of this helps
If it does, gimme some Kudos

Cheers!
J
Re: A set consist of 2n-1 element from Walker's collection   [#permalink] 29 Dec 2013, 17:31
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